Human Eye and Colourful World PDF

CHAPTER Human Eye and 2 Colourful World The Human Eye Human eye is the most delicate and complicated natural optical instrument. It is used to see the beautiful nature and the natural phenomena. It is a spherical ball of diameter...

CHAPTER Human Eye and 2 Colourful World The Human Eye Human eye is the most delicate and complicated natural optical instrument. It is used to see the beautiful nature and the natural phenomena. It is a spherical ball of diameter about 2.3 cm. Its essential parts are described below : zz Sclerotic : It is the outermost coating of the eye ball. It is tough, white and opaque and forms white of the eye. It keeps eye ball in spherical shape and protects it from shocks and injury. zz Cornea : The transparent front projected part of eye ball is called cornea. It is a thin, transparent and white bulging portion of sclerotic in the front. It refracts most of the light rays into the eye. zz Choroid : It starts below the sclerotic. It is grey-black layer of tissues and its function is to keep the eye dark by absorbing diffused light falling on it and to avoid internal reflections within the eye. zz Iris : It is a dark muscular diaphragm that controls light level inside the eye. zz Pupil : It is the central circular aperture in the iris. Its normal diameter is 1 mm but it can contract in excess light and expand in dim light, by means of two sets of involuntary muscular fibres. zz Crystalline lens : The crystalline lens merely provides the finer adjustment of focal length required to focus objects at different distance on the retina. zz Ciliary muscles : Ciliary muscles controls the focal length of the eye lens. When these muscles contracts, the focal length of the lens increases. On the other hand, when they expand, they exert inward pressure on the eye lens and decrease the focal length of the lens. zz Optic nerve : It is a bundle of nerves originating from the brain and entering the eye ball from the posterior side. It attaches itself to the choroids. Its function is to carry electrical signals to the brain. 50 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics zz Retina : It forms innermost coat in the interior of the eye. It consists of a thin membrane which is rich in nerve fibres, containing two kinds of vision cells (rods and cones) and blood vessels. It is sensitive to light, for it is a continuation of the optic nerves. It serves the purpose of a sensitive screen for the reception of the image formed by the lens system of the eye. The retina possesses following two important spots : zz Yellow spot : It is situated at the centre of the retina. It is a slightly raised spot with a minute depression in its peak. It is yellow in colour and most sensitive to light. The central region of the yellow spot is called the fovea centalis. zz Blind spot : The point where optic nerve enters the eye ball is called blind spot. It has no nerve endings. It is the area on the retina which does not have light sensitive cells (rods and cones). So if the image is formed on this part, then no signal is sent to the brain. As a result object is not visible to the eye. 1 The chick wakes up late in the morning and goes to sleep early the evening why? Ans.: The retina of chick's eye consists of very few rod shaped cells. Due to this it can see only Ans.: The retina of bees contains cone cells which in bright light. are sensitive bee ultra-violet light. Human beings do not have those cone cells and therefore, we 2 Bees are able to see ultra-violet light but are ultra-violet blind. human beings cannot. Explain. There are 100 million rod cells and 10 million cone cells on the Do You retina. Some animals have their two eyes on the opposite sides Know of head, i.e., one in the front and other at the back of head. This gives a much wider field of view. Accommodation of an Eye A normal eye can see the near and far off objects clearly if the sharp images of these objects are formed on the retina. Since the distance between the eye lens and retina is fixed (i.e., v = constant), so to see the objects at different positions from the eye lens, the focal length of the lens has to be changed accordingly  1 1 1 to form the sharp images of these objects  ∵ − + = .  u v f If the object to be seen is far off (i.e., at infinity), then the sharp image of this object can be formed on the retina of the eye by increasing the focal length of the eye lens. The focal length of a lens increases if its thickness decreases. Thus, to decrease the thickness of the eye lens and hence to increase its focal length, ciliary muscles are completely relaxed. Now the parallel beam of light coming from the far off object is focused on the retina and hence the object is seen clearly. Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 51 If the object to be seen is close to the eye i.e., near to the eye, then the sharp image of this object can be formed on the retina of the eye by decreasing the focal length of the eye lens. The focal length of a lens decreases if its thickness increases. Thus, to increase the thickness of the eye lens and hence to decrease its focal length, ciliary muscles contract. Now the beam of light coming from the near object is focused by the lens on the retina. Hence the object is seen clearly. Conclusion Thus, we observe that the focal length of the eye lens is adjusted automatically by the action of ciliary muscles such that a sharp images of the objects at different positions from the eye are formed on the retina. This process is known as accommodation of an eye. The ability of an eye to change the focal length of its lens in such a way that the sharp image of an object at different positions from the eye is formed on the retina is known as accommodation of an eye. Do You An electron microscope uses an electron beam in place of light to see images of very small objects. It has a magnifying power of more Know than 5,00,000. It is used to see images of atoms and molecules. Near Point The nearest point up to which the eye can see clearly is called the near point of the eye. For a young adult with normal vision, the near point is about 25 cm. This distance is known as least distance of distinct vision. Far Point The most distant object, the eye can see is called the far point of the eye. For a normal eye it is at infinity. A normal eye can focus the objects situated anywhere from 25 cm to infinite distances. Range of Vision The distance between the near point and the far point is called range of vision of the eye. Within the range of vision, there is one point where object placed are most distinctly visible. The distance of this point, from the eye, is called least distance of distinct vision. For normal eye this least distance is 25 cm. Power of Accommodation The maximum variation in power of the lens to see nearby objects and distant objects clearly is called power of accommodation. Power of accommodation of eye is the ability of the eye to observe distinctly the objects situated at widely different distances from the eye, on account of change in focal length of eye lens by the action of ciliary muscles holding the lens. The maximum power of accommodation of the eye for a person 100 100 100 having normal vision (d = 25 cm) is P = = = = 4 dioptre. f d 25 Do You Focal length of an eye lens cannot be decreased below a Know certain limit. 52 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics Persistence of Vision It is one of the important characteristics of human eye. The image of an object formed on the retina of the eye is neither permanent nor it fades away instantly. In fact, the image of any object seen persists on the retina for 1/16 second, even after the removal of the object. This continuance of sensation of eye for some time is called persistence of vision. If another object is seen before the impression of the first object fades away completely, the eye is not able to separate the two impressions, and a sense of continuity developes. To study the 3D effect. This simple demonstration shows exactly how the eyes are used during parallel-viewing. For this pick a specific object at a certain distance. Aim your eyes at that target. While looking at that distant target, bring your index fingers, tips touching, up in front of your eyes and in to your line of sight. While still aiming your eyes at the distant target, calmly notice that a mini-frank has appeared between the tips of your fingers. Do not allow the beauty of the mini-frank to distract you at it. Continue to aim your eyes into the distance at your target. Pull the tips of your finger apart slightly and observe the frank furter floating in the air. Wiggle your fingers and watch the mini-frank dance. Remember how your eyes feel while performing this feat and you can apply the same skills to 3D viewing. Defects of Vision and their Correction Sometimes, the eye may gradually lose its power of accommodation. In such conditions, the person cannot see the objects distinctly and comfortably. The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are : zz Myopia (near-sightedness or short sightedness) zz Hypermetropia (far-sightedness or long sightedness) zz Presbyopia Myopia (Near-sightedness or Short sightedness) Myopia is that defect of human eye by virtue of which it can see clearly the objects lying at short distance from it. But the far off objects cannot be seen clearly by the myopic eye. Causes of Myopia : The two possible causes of this defect are : zz Increase in the length of the eye ball, as if distance of retina from the eye lens has increased zz Decrease in focal length of the eye lens when the eye is fully relaxed. This is as if the ciliary muscles holding the eye lens do not relax fully and have some tension. Correction : The image of a distant object (i.e., at infinity) is formed in front of the retina of eye suffering Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 53 from myopia as shown in figure. (a) As the image of the object lying at infinity is not formed on the retina of the eye, so such object can not be seen clearly by the myopic eye. The far point of such an eye is near to the eye as shown in figure (b). This defect can be corrected by using a concave lens of suitable focal length. So, a man suffering from this defect wears spectacles having concave lens of suitable focal length. The concave lens diverges the rays of light entering the eye from infinity. Hence this lens makes the rays of light appear come from the far point (O′) of the defective eye as shown in figure (c). Let x = distance of far point of myopic eye, f = focal length of concave lens to be used. As the object to be seen is at infinity and its image is to be formed at the far point, therefore, u = ∞ and v = – x. Distance of far point O′ from eye lens is taken same as the distance of far point O′ from concave lens. 1 1 1 From, = − or f = – x f v u Hence, focal length of concave lens used for correcting the myopic eye is equal to distance of far point of the myopic eye. 3 The far point of a myopic person is 80 cm. What is the focal length of the lens required to enable him to see very distant objects clearly? 4 The above person prefers to remove his An s.: F ar p oi n t of n orma l h uma n e ye i s spectacles while reading a book. Explain why? infinity. i.e., Ans.: The myopic person may have a normal near u = – ∞ point, i.e. about 25 cm (or even less). In order to and v = – 80 cm read a book with the spectacles, this person must Use lens formula: keep the book at a distance greater than 25 cm so that the image of the book by the concave lens is 1 1 1 − = produced not closer than 25 cm. The angular size v u f of the book (or its image) at the greater distance 1 1 1 is evidently less than the angular size when the − = − 80 − ∞ f book is placed at 25 cm and spectacles are not f = –80 needed. Hence, the person prefers to remove the negative focal length means concave lens. spectacles while reading. 54 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics Hypermetropia (Far-sightedness or Long sightedness) A human eye which can see far off objects or distant objects clearly but can not see the near objects clearly is said to be suffered with a defect known as long sightedness or far sightedness or Hypermetropia. Causes : The two possible causes of this defect are : zz Decrease in length of the eye ball, as if distance of retina from the eye lens has decreased. zz Increase in focal length of the eye lens, when the eye is fully relaxed. Correction : The image of a normal near point (which is 25 cm from the eye lens) is formed behind the retina of eye having long-sight defect as shown in figure (a). The image of the normal near point formed on the retina is blurred. The near point of such eye is little far from the near point of normal eye as shown in figure (b). This defect can be corrected by using a convex lens of suitable focal length. So, a man suffering from this defect wears spectacles having convex lens of suitable focal length. The convex lens of spectacles reduces the divergence of rays of light entering the eye. Hence this lens makes the rays of light appear to come from the near point of the defective eye as shown in figure (c). Let x = distance of near point O’ of defective eye, d = distance of near point O of normal eye = least distance of distinct vision of normal eye. f = focal length of convex lens to be used. For the correcting lens, object is at O, i.e., u = – d Image is at O’, i.e., v = – x 1 1 1 1 1 1 −d + x xd As = − , ∴ = + = ⇒ f= f v u f −x d xd x−d As x > d, f is positive. Therefore, correcting lens must be convex. 5 A person cannot see distinctly any object placed beyond 40 cm from his eye. Calculate the power of the lens which will enable him to see is infinity. So he should use lens which forms distant stars clearly. image of distant object (u = – ∞) at a distance of Ans.: As the person cannot see objects clearly 40 cm is in front of it. beyond 0.4 m, his far point is 0.4 m. He wants 1 1 1 10 to see distant stars clearly, i.e., distance of object − − = = P i.e., P = = − 2.5 D 0.40 − ∞ f −4 Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 55 Presbyopia A human eye which cannot see the near objects clearly is said to suffer from a defect known as presbyopia. Eye suffering from presbyopia cannot read and write comfortably. This is a kind of hypermetropia which is experienced in old age and therefore, also called old age hypermetropia. Cause of Presbyopia : This defect arises due to the ageing of a person. The ciliary muscles are weakened and the flexibility of the crystalline lens of the human eye decreases with age of the person. As a result, human eye is unable to focus on close objects. Correction of presbyopia : (i) This is corrected by using a convex lens of suitable power, which is added to glasses, if any for the distant vision. (ii) Sometimes a person may suffer from both the defects of vision, i.e., myopia and hypermetropia. Such type of defects can be corrected by using bifocal lenses which consist of both concave and convex lenses. The upper portion of the spectacles consists of concave lens for the person to see distant objects and the lower portion consists of convex lens which help him to see nearby objects. To correct presbyopia, an old person has to use spectacles with a convex lens of suitable focal length, or power. The cause of hypermetropia is decrease in length of eye ball or increase in focal length of eye lens. But the cause of presbyopia is only increase in focal length of eye lens. The eye ball, in presbyopia, has normal length. zz Presbyopia is a natural part of ageing process. It is not a Do You disease and cannot be prevented. Know zz An eye can suffer from myopia, hypermetropia as well as astigmatism simultaneously. Prism Prism is a homogenous transparent refracting medium bounded by at least two non-parallel surfaces inclined at some angle. The two plane surfaces PQBD and PRCD are called as refracting faces. 56 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics The line along which the two refracting faces meet is called refracting edge of the prism. The angle between the two refracting faces is called the angle of prism and is denoted by A. The section DBC of the prism is known as principal section. Refraction of Light through a Prism Figure shows the principal section of a prism DBC with angle A. Consider a ray of light PQ incident on refracting face DB at point Q. N1N is the normal at point Q. The incident ray PQ is refracted along QL and bends towards normal. The refracted ray QL again suffers refraction at L and emerges out along LM away from normal N2N at point L. The ray PQ, QL and LM are known as incident ray, refracted ray and emergent ray respectively. Further, the angles PQN1 = ∠i1, MLN2 = ∠i2, LQN = ∠r1 and QLN = ∠r2 are known as angle of incidence, angle of emergence, angle of refraction at surface DB and angle of incidence at surface DC respectively. If the incident ray is produced forward and emergent ray LM is produced backward, then the two rays meet at O. The angle ∠TOL = d is known as angle of deviation. So, the angle of deviation means the angle between emergent and incident rays, i.e., the angle through which the incident ray has turned in passing through the prism. From triangle QOL, d = (i1 – r1) + (i2 – r2) = (i1 + i2) – (r1 + r2) … (i) From quadrilateral DQNL, A + 90° + q + 90° = 360° or A + q = 180° … (ii) From triangle QNL, r1 + r2 + q = 180° … (iii) From eqs. (ii) and (iii), r1 + r2 + q = A + q \ r1 + r2 = A … (iv) Substituting the value in eq. (i) we get d = (i1 + i2) – A or i1 + i2 = d + A … (v) To find angle of deviation of a ray of light passing through the prism. Procedure : Place a glass prism on a sheet of white paper. Mark its boundary with a pencil. Now fix two pins at points B and C as shown in figure. Look through the refracting face ac and fix two pins at H and I such that the pins H and I and the images of pins B and C line in a straight lie. Now remove the prism and all pins. Join the points B and C and extend this line so that it touches the refracting surface ab at point E. Also join the points H and I and extend this line so that it touches the refracting surface ac at point G. Also join points E and G. Draw perpendiculars N1N′1 and N2N′2 on the refracting surface ab at point E and on the refracting surface ac at point G respectively. ∠N1EC = ∠i, angle of incident and the ray BCE is the incident ray. ∠N2GH = ∠i2, angle of emergence and the ray GHI is the emergent ray. Produce the emergent ray GHI backward so that it intersects the incident ray produced forward at point J. ∠FJG = ∠D, angle of deviation of the incident ray of light passing through the prism. Thus, angle of deviation of a ray of light is defined as the angle between the directions of incident ray of light and the emergent ray of light. Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 57 Dispersion of White Light by a Glass Prism When a ray of white light (sunlight) enters a glass prism (denser medium), it emerges out from it, broken into seven colours. This phenomenon, due to which different components of a white light are separated by a denser medium, is called dispersion. The phenomenon of splitting of white light into seven colours when it passes through a glass prism is called dispersion of white light. When the dispersed white light is made to fall on a white screen, we get a seven coloured band of light. This coloured band is called spectrum. The order of the colours from the lower end of the spectrum is violet, indigo, blue, green, yellow, orange and red. The colours can be remembered by the word VIBGYOR. In air, light waves of all colours have same velocity (3 × 108 m s–1). White light is a mixture of seven colours. Every colours has its own characteristic wavelength. Wavelength of different colours are given in the table. S. NO. Colours Wavelength 1 Violet 4000 Å 2 Indigo 4500 Å 3 Blue 4800 Å 4 Green 5400 Å 5 Yellow 5800 Å 6 Orange 6000 Å 7 Red 7900 Å Recombination of colours of spectrum to give white light Isaac Newton was the first to use a glass prism to obtain the spectrum of sunlight. He tried to split the colours of the spectrum of white light further by using another similar prism. However, he could not get any more colours. He then placed a second identical prism in an inverted position with respect to the first prism, as shown in figure. 58 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics This allowed all the colours of the spectrum to pass through the second prism. He found a beam of white light emerging from the other side of the second prism. This observation gave Newton the idea that the sunlight is made up of seven colours. Any light that gives a spectrum similar to that of sunlight is often referred to as white light. The first prism is known as dispersing prism as dispersion takes place at the first face of prism (A) and second prism (B) recombines the dispersed colours therefore it is known as recombination prism. Both the prism A and B together act as a glass slab with parallel sides, and emergent ray is parallel to the incident ray. Do You In vacuum, dispersion of light does not take place because all Know colours travel with same speed in vacuum. 6 What will be the colour of the emergent light when white light is incident on a thin walled hollow glass prism. Ans.: The emergent light is white because the outer faces of the prism behave like hollow plates as shown in figure. To show that a glass prism splits colours of white light passing through it and does not produce any colour by itself. After obtaining the spectrum of sunlight; Newton tried a few more experiments. He tried to further split up the colours of the spectrum by using another similar prism. For this he made a small hole in the white screen and obtained the spectrum from the first prism on it. He adjusted the screen in such a way that only a limited portion of spectrum (a single colour say yellow colour) passes through the hole. This is then made to fall on the similar prism B which is placed in an inverted position with respect to the screen. This shows that prism does not produce any colour by itself. Dispersion takes place at first face of the prism and afterwards at second prism it gets refracted only. Dispersion of Light in Nature When sunlight falls on the water drops suspended in the atmosphere after rainfall, rainbow is formed due to the dispersion of sun light. The water drops suspended in air (or atmosphere) act as prisms. Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 59 Rainbow It is the example of dispersion of sunlight. When sunlight falls on a water drop suspended in air, then the sunlight is refracted. The refracted sunlight splits (or dispersed) into its constituent colours (i.e., seven colours). Thus, water drop suspended in air behaves as a glass prism. The red colour deviates the least and the violet colour deviated the most. Different colours of refracted sunlight fall on the opposite face of the water drop. Now, each colour is partly reflected back into the drop. The reflected colours on reaching the lower surface of water drop are refracted again into the air. Thus, we get a spectrum of seven colours, which is known as a rainbow. Rainbow is observed during a rainfall or after the rainfall or when we look at a water fountain provided the sun is at the back of the observe. Atmospheric Refraction When sunlight enters the earth’s atmosphere, then it continuously goes from rarer to the denser medium therefore, refraction of light takes place. The refraction of light taking place in the atmosphere is known as atmospheric refraction. Twinkling of Stars Light emitted by stars passes through the atmosphere of the earth before reaching our eyes. The atmosphere of the earth is not uniform but consists of many layers of different densities. The layers close to the surface of the earth are optically denser. As we go higher and higher, the density of layers and refractive index decreases progressively. As the light from a star enters the upper-most layer of the atmosphere, it bends towards the normal as it enters the next layer. This process continues till the light enters our eyes. So due to refraction of light, the apparent position of the star is different from the actual position of the star. Moreover, the different layers of the atmosphere are mobile and the temperature and the density of layers of atmosphere changes continuously. Hence, the apparent position of the star changes continuously. The change in the apparent position of the star continuously leads to the twinkling of a star. 60 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics  Planets do not twinkle, because planets are very close to Do You the earth as compared to the stars. So the intensity of light Know we receive from the planets is very large. Therefore, the variation in the brightness of the planets is not detected. Advanced Sunrise and Delayed Sunset The sun is visible to us about 2 minutes before the actual sunrise, and about 2 minutes after the actual sunset because of atmospheric refraction. By actual sunrise. We mean the actual crossing of the horizon by the sun. In the figure the actual and apparent positions of the sun with respect to the horizon is shown. The time difference between actual sunset and the apparent sunset is about 2 minutes. The apparent flattening of the Sun’s disc at sunrise and sunset is also due to the same phenomenon. Thus, we gain about 4 minutes of additional daylight each day. The Sun Appears oval (flattened) at Sunset and Sunrise but Appears Circular at Noon At sunset and sunrise, the sun is near the horizon. The rays of light from the upper and lower edge of the sun bend unequally while traveling through the atmosphere. As a result of this phenomenon, the sun appears oval or flattened. At noon, the sun is overhead. The rays of light from the sun enter the atmosphere normally and hence they do not bend at all while passing through the atmosphere. Therefore, the sun appears circular at noon. Scattering of Light When sunlight enters the atmosphere of the earth, the atoms and molecules of different gases present in the atmosphere absorb this light. Then these atoms and molecules of the gases re-emit light in all directions. This process is known as scattering of light. We know that light travels in the form of electromagnetic radiations, when these radiations interact with matter, scattering of light takes place. This phenomenon is exhibited practically by all matter. It can even be observed when a strong beam of light is passed through colloidal solution; the path of beam becomes visible. Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 61 Another important factor in scattering of light is relative size of the scatterers (say x) compared to the wavelength of light (l). For x < < l, i.e., when size of scatterer is much less than the wavelength of light, Rayleigh scattering is valid. 1 According to Rayleigh scattering, intensity of light scattered ∝. λ4 However, when x >> l, i.e., size of scatterer is much larger than the wavelength of light, Rayleigh scattering is not valid. In that event, all wavelengths are scattered nearly equal. These conclusions regarding the scattering of light have been verified experimentally by Tyndall. Tyndall Effect The phenomenon of scattering of light by dust, smoke and tiny water droplets suspended in air is known as Tyndall effect. Examples of Tyndall effect are (i) When a beam of sunlight enters a dusty room through a window, then path becomes visible to us. (ii) A beam of sunlight becomes visible as it passes through dust particles in the air of a room. (iii) Tyndall effect is seen, when sun light comes down through the clouds. (iv) When sunlight passes through a canopy of a dense forest, tiny water droplets in the mist scatter light. Do You Coloured television sets and computer monitors produce the entire range of colours by using red, green and blue Know phosphors. To study the scattering of light in colloidal solution. Place a strong source (S) of white light at the focus of a converging lens (L1). This lens provides a parallel beam of light. Allow the light beam to pass through a transparent glass tank (T) containing clear water. Allow the beam of light to pass through a circular hole (C) made in a cardboard. Obtain a sharp image of the circular hole on a screen (MN) using a second converging lens (L2), as shown in Figure. Dissolve about 200 g of sodium thiosulphate (hypo) in about 2 L of clean water taken in the tank. Add about 1 to 2 mL of concentrated sulphuric acid to the water. What do you observe? You will find fine microscopic sulphur particles precipitating in about 2 to 3 minutes. As the sulphur particles begin to form, you can observe the blue light from the three sides of the glass tank. This is due to scattering of short wavelengths by minute colloidal sulphur particles. Observe the colour of the transmitted light from the fourth side of the glass tank facing the circular hole. It is interesting to observe at first the orange red colour and then bright crimson red colour on the screen. 62 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics Applications of Scattering of Light The blue colour of the clear sky In white light, the blue colour has smallest wavelength and in the atmosphere the molecules of air and other fine particles have sizes smaller than the wavelength of visible light. Therefore blue light is scattered more than any other colour. That is why sky appears blue. If earth had no atmosphere there would not have been any scattering and the sky would have looked dark. That is why it appears dark to the passengers flying at very high altitudes, as scattering is not prominent at such heights. Colour of the sun at sunrise and sunset At the time of sunrise and sunset, the position of the sun is very far away from us. The sunlight travels longer distance through the atmosphere of the earth before reaching our eyes. Scattering of blue light is more than the scattering of red light. As a result of this, more red light reaches our eyes than any other colour. Therefore at sunset and sunrise sun appears red. When the sun is overhead in the noon, the light waves of shorter wavelengths of light-blue and violet get scattered by the particles near the horizon but the longer wavelengths (colours orange and red) are scattered least, i.e., they travel relatively undisturbed and reach the earth. White colour of clouds We know that if the size of the scattering particles is large enough, than the scattered light may even appear white. Clouds are generally at lower heights of the atmosphere and lower atmosphere contains large dust particles and water droplets which scatter all colours or wavelength and clouds generally appear white. Danger signals are of red colour When light falls on the signal, all colours are scattered much more than that of red colour. So the red colour suffering least scattering remains confined around the signal, which in turn illuminates the signal significantly. Thus, the danger signals can be seen from very far off distances. Moreover, among all colours, red colour or red light is scattered least by smoke or fog. Hence red signals are visible even through the smoke or fog. Yellow colour headlights in foggy weather In a foggy weather scattering of light is large because of more particles (scatters) are present in air. Yellow colour light is used in the headlights of the vehicles because wavelength of yellow colour of the light is large and it does not undergo large scattering. It is observable from a large distance and is easy to produce. Sky appears greyish over cities having industrial units. Do You The smoke and dust particles in the atmosphere over such cities scatter red, orange and yellow colours more than Know other colours of small wavelength. Hence, the sky appears greyish. Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 63  CONCEPT MAP 64 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics zz Human Eye : It is a light sensitive organ which enable us to see. zz Parts of human eye are cornea, iris, lens, ciliary muscles, retina and optic nerve. zz Cornea is the aperture or window of an eye. It allows light to enter in the eye. zz Iris controls the amount of light entering the eye through pupil. zz Ciliary muscles : It increases or decreases the focal length of the eye lens. zz Retina : It acts as the screen or film to obtain the image of an object. zz Optic nerve : It carries signal to the brain for interpretation. zz Accommodation of an eye : It is the ability of eye lens to change its focal length to form sharp images of objects at different positions from the eye on the retina of the eye. zz Near point : It is the nearest position of an object from human eye so that its sharp image is formed on the retina. zz Near point of a normal human eye is 25 cm. zz Far point is the farthest position of an object from a human eye so that the sharp image of the object is formed on the retina. zz Range of vision : It is the distance between the near point and the far point of an eye. zz For normal human eye, the range of vision is 25 cm to infinity zz Power of accommodation of a normal human eye is about 4 dioptre. zz Myopia or near-sightedness : A person with myopia can see nearby objects clearly but cannot see distant objects distinctly. It is caused by (i) excessive curvature of the eye ball (ii) elongation or increase in the size of the eye lens. It can be corrected by using spectacles with a concave lens of suitable focal length. zz Hypermetropia or far-sightedness: A person with hypermetropia can see distant objects clearly but cannot see nearby objects distinctly. It is caused by (i) increase in the focal length of the eye lens (ii) shortening or decrease in the size of the eye lens. It is corrected by using spectacles made from convex lens of suitable focal length. zz Presbyopia : In this defect, the power of accommodation of the eye decreases with ageing. It can be corrected by using a bifocal lens consisting of a concave and convex lens. zz Astigmatism: A human eye which cannot focus on both horizontal and vertical line simultaneously suffers from astigmatism. zz Cataract: An eye in which an opaque membrane is formed over the eye lens suffers form cataract. Cataract can be corrected by performing surgery. zz Prism : It is a homogeneous transparent refracting medium bounded by at least two non-parallel surfaces inclined at some angle. zz Dispersion of white light: The phenomenon of splitting of white light in to seven colours when it passes through a glass prism is called dispersion of white light. zz Red colour deviates the least while passing through a glass prism. zz Violet colour deviates the most while passing through a glass prism Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 65 zz Spectrum: The band of colour components of a light beam obtained on a white screen when white light passes through a prism is called the spectrum. zz Rainbow: A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on raindrops after the rain. The raindrops act like small prisms. zz Atmospheric refraction: When the light rays pass through the atmosphere having layers of different densities and refractive indices, then refraction of light takes place. This refraction of light by the earth’s atmosphere is called atmospheric refraction. zz Scattering of light: It is a phenomenon of change in the direction of light on striking a particle like an atom, a molecule, dust particles, water droplets, etc. It explains the phenomenon like blue colour of the sky, the reddening of the sun at sunrise and the sunset, etc. 1 zz Intensity of scattered light, I ∝ , if size of scatterer is small as compared to the wavelength (l) of light. λ4 zz In a clear atmosphere of the earth, colours of small wavelengths like violet, and blue are scattered more than red colour. zz In a polluted atmosphere of the earth, the scattering of colours of higher wavelengths is more than the scattering of colours of smaller wavelengths. zz Blue colour of sky, greenish-blue colour of sea water, red colour of sunset and sunrise and white colour of clouds are due to scattering of light. 66 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics  SOLVED EXAMPLES 1. Calculate the power of the eye lens of the defects and to increase his range or vision normal eye when it is focused at its (a) far from 25 cm to infinity. In a bi-focal lens, point, infinity and (b) near point, 25 cm from upper portion is concave which corrects the eye. Assume the distance of the retina distant vision and lower portion is convex from the eye lens to be 2.5 cm. which corrects near vision. Ans.: (a) When the object is at infinity, the image 3. When white light passes through a glass forms at the focus of the lens (v = f ). prism, seven colours namely red, orange, Hence, the focal length in this case is yellow, green, blue, indigo and violet are 2.5 cm. Thus the power is seen on the white screen. All these colours 1 1 have different angles of deviation. Explain P1 = = = 40 D f 2.5 × 10 −2 m why? (b) In this case, the object is at 25 cm from Ans.: Speed of colour in a medium depends upon the eye lens, and the image is formed its wavelength. All the colours have different at 2.5 cm from the eye lens. wavelength. The red colour has the longest So, u = – 25 cm, v = 2.5 cm. Then, wavelength and violet colour has the least 1 1 1 1 1 wavelength. Therefore, red colour has the = − = − highest speed in the glass prism and the f v u 2.5 cm 25 cm violet colour has the lowest speed in the glass 100 100 \ P2 = + = 44 m–1 = 44 D prism. Hence, all colours of white light are 2.5 m 25 m refracted by different amount while passing The maximum variation in the power through the glass prism. Therefore, all the is, 44 D – 40 D = 4 D colours have different angles of deviations. 2. A person is able to see object clearly only 4. The near point of an elderly person in 50 when these are lying at distance between cm from the eye. Find the focal length and 50 cm and 300 cm from his eye. power of the corrective lens that will correct (i) What kind of defect of vision he is his vision. suffering from? Ans.: The near point of an elderly person = 50 cm (ii) What kind of lenses will be required from the eye. to increase his range of vision from Hence, v = – 50 cm, u = – 25 cm 25 cm to infinity? Explain briefly. 1 1 1 −1 1 −1 + 2 1 Ans.: (i) For a normal eye, the near point is at = − = + = = f v u 50 25 50 50 cm 25 cm and the far point is at infinity from the eye. The given person cannot see \ f = 50 cm or 0.5 m object clearly either closer to the eye or 1 1 Power of the lens = = = + 2 dioptres far away from the eye. So, he is suffering f 0.50 m from both myopia and hypermetropia. 5. A person cannot see distinctly any object (ii) A bi-focal lens consisting of a concave placed beyond 40 cm from his eye. Calculate lens and convex lens of suitable focal the power of the lens which will enable him lengths will be required to correct the to see distant stars clearly. Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 67 Ans.: As the person cannot see objects clearly 8. Why the sun appears oval at sunrise and beyond 0.4 m, his far point is 0.4 m. He wants sunset, and circular at noon? to see distant stars clearly, i.e., distance of Ans.: This can be explained in terms of atmospheric object is infinity. So he should use lens which refraction. At sunrise and sunset, the sun is forms image of distant object (u = – ∞) at a near the horizon. The rays of light from the distance of 40 cm in front of it. upper part and lower part of the periphery 1 1 1 of the sun bend unequally on travelling Using lens formula, − = v u f through earth’s atmosphere. That is why the 1 1 1 sun appears oval or flattened at the time of − = =P −0.40 −∞ f sunrise and sunset. 10 At noon, the sun is overhead. The rays of light i.e., P=− = − 2.5 D 4 from the sun enter earth’s atmosphere normally. 6. (a) What is the power of the eye glasses Therefore, they suffer no refraction or bending worn by a person whose far point is on passing through earth’s atmosphere. Hence 5 m? the sun appears circular at noon. (b) Locate the virtual image of an object 9. What is meant by accommodation of human 2 m in front of the eye glasses. eye? How is it achieved? Ans.: (a) As here far point is 5 m while for normal Ans.: Accommodation or power of accommodation eye it is ∞, of human eye is the ability of the eye to u = – ∞ and v = – 5 m observe distinctly the objects situated at widely 1 1 1 So, − = = P i.e., P = – 0.2 D different distances from the eye. This property −5 −∞ f is due to the action of ciliary muscles holding (b) If the object is at a distance of 2 m the eye lens. For observing distant objects, 1 1 1 1 1 1 eye is in relaxed state, i.e., unaccommodated. × = i.e., = − v 2 −5 v −5 2 The eye lens is thin and image of distant 1 7 10 object is formed on the retina. i.e., =− or v = = 1.43 m v 10 7 For observing nearby objects, the ciliary i.e., the virtual image of object which is muscles contract, the eye is in a state of at 2 m will be formed at a distance of tension. It is said to be accommodated. 1.43 m from the lens on the same side. 10. What is meant by range of vision? What is 7. A person needs a lens of power –6.5 dioptres its value for a person with normal vision? for correcting his distant vision. For correcting Ans.: The range of vision of a person is the his near vision he needs a lens of power distance between near point (N) and far + 2.5 dioptre. What is the focal length of point (F) of his eyes. The point at closest the lens required for correcting (i) distant distance, at which an object can be seen vision, and (ii) near vision? clearly by the eye is called near point (N) Ans.: (i) Focal length of distance vision of the eye. 1 −100 The most distant point at which an object = = cm = − 15.38 cm Power 6.5 can be seen clearly is called far point (F) of (ii) Focal length in near vision the eye. For a normal eye, far point lies at 100 infinity. Therefore, for a person with normal = cm = 40 cm 2.5 vision, the range of vision is infinite. 68 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics  NCERT SECTION 1. What is meant by power of accommodation (c) accommodation of the eye? (d) far-sightedness Ans.: Power of accommodation of eye is the ability Ans.: (c) The property of the eye to adjust of the eye to observe distinctly the objects the focal length of eye lens is called situated at widely different distances from accommodation. Choice (c) is correct. the eye, on account of change in focal length 6. The human eye forms the image of an object of eye lens by the action of ciliary muscles at its holding the lens. (a) cornea (b) pupil 2. A person with a myopic eye cannot see (c) iris (d) retina objects beyond 1.2 m distinctly. What should Ans.: (d) The human eye forms the image of an be the type of the corrective lens used to object at its retina. restore proper vision? 7. The least distance of distinct vision for a Ans.: The corrective lens should form the image of young adult with normal vision is about the far off object at the far point of the myopic (a) 25 m (b) 2.5 cm person. So, by using the lens formula, we can (c) 25 cm (d) 2.5 m write Ans.: (c) The least distance of distinct vision for 1 1 1 1 1 1 a young adult with normal vision is = − = − =− 25 cm. f v u −1.2 m ∞ 1.2 m or, f = – 1.2 m 8. The change in focal length of an eye lens 1 is caused by the action of So, power of the lens, P = − D = − 0.83 D (a) pupil (b) retina 1.2 (c) iris (d) ciliary muscles 3. What is the far point and near point of the Ans.: (d) The action of ciliary muscles holding human eye with normal vision? the eye lens changes the focal length of Ans.: The nearest point up to which an eye can eye lens enabling the eye to focus the see clearly is called its near point. For a image of objects at varying distances. normal eye, the near point is at a distance of 25 cm. 9. A person needs a lens of power –5.5 dioptres The farthest point up to which an eye can see for correcting his distant vision. For correcting clearly is called its far point. For a normal his near vision, he needs a lens of power eye, the far point is at infinity. + 1.5 dioptre. What is the focal length of the lens required for correcting (a) distant 4. A student has difficulty in reading the vision, and (b) near vision? blackboard while sitting in the last row. What Ans.: For the near-sighted person, the power of the could be the defect the child is suffering lens used for distant-viewing is –5.5 D. from? How can it be corrected? So, For distance viewing Ans.: The student is suffering from myopia. Myopia 1 1 1 = = = can be corrected by using glasses made from Power −5.5 D −5.5 m −1 concave lens of suitable focal length. = – 0.18 m = –18 cm 5. The human eye can focus objects at different Since the near-viewing section of the lens is distances by adjusting the focal length of corrected by + 1.5 D, the eye lens. This is due to \ Focal length of near vision correction, (a) presbyopia 1 1 = = =.67 m = 67 cm (b) near-sightedness Power 1.5 Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 69 10. The far point of a myopic person is 80 cm not be able to see clearly any object placed in front of the eye. What is the nature and closer than 25 cm. power of the lens required to correct the problem? 13. What happens to the image distance in the Ans.: Here, concave lens is required. eye when we increase the distance of an The corrective concave lens should form object from the eye? the image of the far off objects i.e. lying at Ans.: For a normal eye, image distance in the infinity at the far point of the myopic person. eye is fixed and is equal to the distance of So, using the lens formula, retina from the eye lens. When we increase 1 1 1 1 1 1 the distance of the object from the eye, focal = − = − =− length of eye lens is changed on account of f v u − 80 cm ∞ 80 cm accommodating power of the eye so as to So, f = – 80 cm keep image distance constant. 100 So, Power of the lens, P = D = − 1.25 D 14. Why do stars twinkle? − 80 Ans.: Due to wind and convection currents, density 11. Make a diagram to show how hypermetropia of the atmospheric layers keeps on changing. is corrected. The near point of a hypermetropic As a result, the position of a star keeps eye is 1 m. What is the power of the lens fluctuating from its mean position. This required to correct this defect? Assume that the near point of the normal eye is 25 cm. fluctuating image of the stars makes them appear twinkling to the observer. Ans.: The diagram to show the correction of hypermetropia is given here. 15. Explain why the planets do not twinkle. Ans.: Planets do not emit light. However, they become visible due to reflection of light falling on them. The planets are much closer to the earth and thus can be considered as the extended source of light. The fluctuations in the light coming from various points of the planet due to atmospheric refraction get averaged out. As a result, no Given: Near point distance for a hypermetropic twinkling of planets is seen. eye, D = 100 cm Near point distance for a normal eye, D = 25 cm 16. Why does the sun appear reddish early in So, v = – 100 cm, u = – 25 cm the morning? Using the lens formula, Ans.: At sunrise and sunset, the sun is closer to 1 1 1 1 1 the horizon. The sunlight near the horizon = − = − passes through denser layers of the air and f v u −100 cm − 25 cm covers larger distance before reaching our −1 + 4 3 = − = eyes. Most of the blue light gets scattered. 100 cm 100 cm The light that reaches our eyes is of longer 100 cm wavelengths, mainly orange and red. That is So, f= = 33.3 cm and 3 why the sun appears red at the sunrise and 100 at the sunset. P = D = + 3.0 D 33.3 17. Why does the sky appear dark instead of 12. Why is a normal eye not able to see clearly blue to an astronaut? the objects placed closer than 25 cm? Ans.: This is because at such huge heights of the Ans.: The least distance of distinct vision for a astronaut, there is nothing to scatter the normal eye is 25 cm. So, a normal eye will sunlight. Therefore the sky appears dark. 70 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics  EXERCISE Multiple Choice Questions (a) accommodation of eye (b) persistence of vision Level - 1 (c) colour blindness (d) least distance of distinct vision 1. The size of the pupil of the eye is adjusted by 9. Far point of normal human eye is (a) cornea (b) retina (a) 25 cm (b) 25 m (c) iris (d) blind spot (c) 500 m (d) infinity 2. When we enter a cinema hall, we cannot see 10. Near point of normal human eye is properly for a short time. This is because (a) 25 cm (b) 25 mm (a) pupil does not open (c) 25 m (d) not fixed (b) pupil does not close (c) adjustment of the size of pupil takes some 11. When ciliary muscles are relaxed, focal length of time the eye lens is (d) none of these (a) maximum (b) minimum 3. When we go out in bright sunlight, the pupil of the eye (c) neither maximum nor minimum (a) contracts (d) cannot say (b) expands 12. Splitting of white light into seven colours on (c) sometimes contracts and sometimes, passing through a glass prism is called expands (a) reflection (b) refraction (d) neither contracts nor expands (c) scattering (d) dispersion 4. Which cells respond to colour of light? 13. Which of the colours of visible light has minimum (a) Rod shaped cells wavelength? (b) Cone shaped cells (a) Violet (b) Red (c) Both types of cells (c) Yellow (d) Green (d) Neither of the two 14. Which of the colour of visible light has minimum 5. Eye lens is a frequency? (a) double convex lens (a) Violet (b) Red (b) double concave lens (c) Yellow (d) Green (c) plano convex lens (d) plano concave lens 15. Which colour has maximum speed in glass? (a) Violet (b) Red 6. The property of persistence of vision is used in (c) Yellow (d) Green (a) short sightedness (b) long sightedness (c) cinematography (d) colour vision 16. For which colour, refractive index of glass is maximum? 7. Focal length of eye lens is (a) Red (b) Violet (a) fixed (c) Green (d) Yellow (b) variable (c) sometimes fixed and sometimes variable 17. Which colour suffers least deviation on passing (d) cannot say through a prism? (a) Red (b) Violet 8. Variable focal length of eye lens is responsible (c) Indigo (d) Blue for Physics | Class 10 HUMAN EYE AND COLOURFUL WORLD | 71 18. Scattering of light involves 29. Concave lens is used in case of (a) reflection (a) presbyopia (b) astigmatism (b) refraction (c) myopia (d) hypermetropia (c) diffraction (d) change in direction of light 30. Cylindrical lens is used in case of (a) presbyopia (b) astigmatism 19. Blue colour of sky is due to (a) scattering of light (b) reflection of light (c) myopia (d) hypermetropia (c) refraction of light (d) diffraction of light Level - 2 20. Twinkling of stars is due to (a) reflection 31. The intensity of scattered light varies inversely (b) dispersion as nth power of wavelength (l) of incident light (c) atmospheric refraction where (d) none of these (a) n = 2 (b) n = 1 21. Coloured band of light obtained by dispersion of (c) n = 4 (d) n = –4 white light is called 32. Which of the following statements is correct (a) mirage (b) spectrum (c) shadow (d) image about rainbow? (a) In primary rainbow, red colour is on the 22. Twinkling of stars is visible when the stars are (a) any where outside and violet colour is on the inside. (b) no definite situation (b) In primary rainbow, violet colour is on the (c) near the horizon outside and red colour is on the inside. (d) over head (c) Secondary rainbow is brighter than primary 23. The phenomena of mirage is seen due to rainbow. (a) total dispersion (d) In secondary rainbow, light wave suffers one (b) total internal interference total internal reflection before coming out. (c) reflection (d) refraction 33. A fish looking up through the water sees the outside world contained in a circular horizon. If 24. In an eye the focal length of the eye-lens increases 4 when the refractive index of water is and the fish is 3 (a) ciliary muscles contracts 12 cm below the surface, the radius of this circle (b) ciliary muscles expands in cm is (c) iris expands (a) 36 5 (b) 4 5 (d) iris contracts 36 25. Short sightedness is also called (c) 36 7 (d) 7 (a) presbyopia (b) astigmatism (c) myopia (d) hypermetropia 34. Prism angle of a prism is 10°. Their refractive index for red and violet colour is 1.51 and 1.52 26. Long-sightedness is also called (a) presbyopia (b) astigmatism respectively. Then dispersive power will be (c) myopia (d) hypermetropia (a) 0.5 (b) 0.15 (c) 0.019 (d) 0.032 27. Eye defect at old age is called (a) presbyopia (b) astigmatism 35. The far point of a myopic person is 80 cm in front (c) myopia (d) hypermetropia of the eye. What is the power of the lens required 28. Convex lens is used in case of to enable him to see the distant objects clearly? (a) presbyopia (b) astigmatism (a) 1.25 D (b) 1.50 D (c) myopia (d) hypermetropia (c) 1 D (d) 1.75 D 72 | HUMAN EYE AND COLOURFUL WORLD Class 10 | Physics 36. Velocity of light in glass is 2 × 108 m s–1 and in air 44. If the refractive index of a material of equilateral is 3 × 108 m s–1. If the ray passes through glass to prism is 3 , then angle of minimum deviation air, calculate the value of critical angle. of the prism is (a) 40° (b) 42° (a) 30° (b) 45° (c) 45° (d) 50° (c) 60° (d) 75° 37. Which of the following phenomena of light are 45. Prism angle and refractive index for a prism for a involved in the formation of a rainbow? 60° and 1.414. Angle of minimum deviation will (a) Reflection, refraction and dispersion be (b) Refraction, dispersion and total internal (a) 15° (b) 30° reflection (c) 45° (d) 60° (c) Refraction, dispersion and internal reflection 46. A piece of cloth looks red in sun light. It is held (d) Dispersion, scattering and total internal in the blue portion of a solar spectrum, it will reflection appear 38. A concave lens of suitable focal length is used for (a) red (b) black correcting a (c) blue (d) white (a) myopic eye (b) hypermetropic eye 47. A student sitting on the last bench can read the (c) both (a) and (b) (d) neither (a) nor (b) letters written on the blackboard but is not able 39. The broad wavelength range of visible spectrum to read the letters written in his textbook. Which is of the following statements is correct? (a) 4000 – 8000 Å (b) 2000 Å – 4000 Å (a) The near point of his eyes has receded away. (b) The near point of his eyes has come closer to (c) 10000 – 2000 Å (d) none of these him. 40. The amplitude of scattered lightly varies with (c) The far point of his eyes has come closer to wavelength l as him. 1 1 (d) The far point of his eyes has receded away. (a) ∝ (b) ∝ 4 λ λ 48. Which of the following statements is correct 1 regarding the propagation of light of different (c) ∝ (d) ∝ l2 2 colours of white light in air? λ 41. The intensity of scattered light varies with (a) Red light moves fastest. amplitude (a) of scattered light as (b) Blue light moves faster than green light. 1 (c) All the colours of the white light move with (a) ∝ a2 (b) ∝ the same speed. a2 (d) Yellow light moves with the mean speed as 1 (c) ∝ a (d) ∝ that of the red and the violet light. a 49. The refractive index of glass is 3/2. Velocity of 42. For Rayleigh elastic scattering size of scatterer (x) light in glass would be should be (a) 3 × 108 m s–1 (b) 2 × 108 m s–1 (a) x ≈ l (b) x > l 8 –1 (c) 10 m s (d) 1.33 × 108 m s–1 (c) x >> l (d) x